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In a certain Algebra 2 class of 27 students, 18 of them play basketball and 17 of them play baseball. There are 11 students who play both sports. What is the probability that a student chosen randomly from the class plays basketball or baseball? Answer:

In a certain Algebra 22 class of 2727 students, 1818 of them play basketball and 1717 of them play baseball. There are 1111 students who play both sports. What is the probability that a student chosen randomly from the class plays basketball or baseball?\newlineAnswer:

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Q. In a certain Algebra 22 class of 2727 students, 1818 of them play basketball and 1717 of them play baseball. There are 1111 students who play both sports. What is the probability that a student chosen randomly from the class plays basketball or baseball?\newlineAnswer:
  1. Use Inclusion-Exclusion Principle: To find the probability that a student plays basketball or baseball, we need to use the principle of inclusion-exclusion. The formula for the probability of AA or BB is P(A)+P(B)−P(A and B)P(A) + P(B) - P(A \text{ and } B).
  2. Calculate P(Basketball)P(\text{Basketball}): First, we calculate the probability of a student playing basketball. P(Basketball)=Number of students playing basketballTotal number of students.P(\text{Basketball}) = \frac{\text{Number of students playing basketball}}{\text{Total number of students}}.P(Basketball)=1827.P(\text{Basketball}) = \frac{18}{27}.
  3. Calculate P(Baseball)P(\text{Baseball}): Next, we calculate the probability of a student playing baseball. P(Baseball)=Number of students playing baseballTotal number of students.P(\text{Baseball}) = \frac{\text{Number of students playing baseball}}{\text{Total number of students}}.\newlineP(Baseball)=1727.P(\text{Baseball}) = \frac{17}{27}.
  4. Calculate P(Basketball and Baseball)P(\text{Basketball and Baseball}): Then, we calculate the probability of a student playing both basketball and baseball. P(Basketball and Baseball)=Number of students playing both sportsTotal number of students.P(\text{Basketball and Baseball}) = \frac{\text{Number of students playing both sports}}{\text{Total number of students}}.P(Basketball and Baseball)=1127.P(\text{Basketball and Baseball}) = \frac{11}{27}.
  5. Apply Inclusion-Exclusion Principle: Now, we apply the principle of inclusion-exclusion to find the probability of a student playing basketball or baseball.\newlineP(Basketball or Baseball)=P(Basketball)+P(Baseball)−P(Basketball and Baseball)P(\text{Basketball or Baseball}) = P(\text{Basketball}) + P(\text{Baseball}) - P(\text{Basketball and Baseball}).\newlineP(Basketball or Baseball)=(1827)+(1727)−(1127)P(\text{Basketball or Baseball}) = (\frac{18}{27}) + (\frac{17}{27}) - (\frac{11}{27}).
  6. Simplify the Expression: We simplify the expression to find the final probability.\newlineP(Basketball or Baseball)=(18+17−11)/27P(\text{Basketball or Baseball}) = (18 + 17 - 11) / 27.\newlineP(Basketball or Baseball)=24/27P(\text{Basketball or Baseball}) = 24 / 27.
  7. Final Probability: Finally, we can simplify the fraction 2427\frac{24}{27} by dividing both the numerator and the denominator by their greatest common divisor, which is 33.
    P(Basketball or Baseball)=(24÷3)(27÷3).P(\text{Basketball or Baseball}) = \frac{(24 \div 3)}{(27 \div 3)}.
    P(Basketball or Baseball)=89.P(\text{Basketball or Baseball}) = \frac{8}{9}.

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